# University of Alberta Guide/STAT/222/Combining Continuous Random Variables

## Convolution

• ${\displaystyle f_{X+Y}(z)=\int _{-\infty }^{\infty }f_{X}(y-z)f_{Y}(z)\delta z=\int _{-\infty }^{\infty }f_{X}(z)f_{Y}(y-z)\delta z}$

### Example

${\displaystyle f_{X}(x)={\begin{cases}{\frac {3x^{2}}{2}}&x\in [-1,1]\\0&else\end{cases}}\qquad f_{Y}(y)={\begin{cases}{\frac {y}{9}}&y\in [4,5]\\0&else\end{cases}}\qquad X{\mbox{ and }}Y{\mbox{ are independent RVs, find }}f_{X+Y}(z)}$

• Start by converting the pdf's to indicator functions
• ${\displaystyle f_{X}(x)={\frac {3x^{2}}{2}}\cdot 1_{[-1,1]}(x)\qquad f_{Y}(y)={\frac {y}{9}}\cdot 1_{[4,5]}(y)}$
• Now ${\displaystyle f_{X}(x)\,}$  is defined only when ${\displaystyle x\in [-1,1]\,}$  and ${\displaystyle f_{Y}(y)\,}$  is defined only when ${\displaystyle y\in [4,5]\,}$
• Use the convolution formula above to write out the integral
• ${\displaystyle \int _{-\infty }^{\infty }f_{X}(z)f_{Y}(y-z)\delta z=\int _{-\infty }^{\infty }{\frac {3z^{2}}{2}}\cdot 1_{[-1,1]}(z){\frac {y-z}{9}}\cdot 1_{[4,5]}(y-z)\delta z}$
• Factor out any constants, in this case, a multiplier
• ${\displaystyle {\frac {3}{18}}\int _{-\infty }^{\infty }z^{2}\cdot 1_{[-1,1]}(z)(y-z)\cdot 1_{[4,5]}(y-z)\delta z={\frac {1}{6}}\int _{-\infty }^{\infty }z^{2}\cdot 1_{[-1,1]}(z)(y-z)\cdot 1_{[4,5]}(y-z)\delta z}$
• Factor out the indicator function for ${\displaystyle (z)\,}$  into the integral bounds
• ${\displaystyle {\frac {1}{6}}\int _{-1}^{1}z^{2}(y-z)\cdot 1_{[4,5]}(y-z)\delta z}$
• Note that ${\displaystyle 4\leq y-z\leq 5,({\mbox{ try to isolate }}z)\rightarrow 4-y\leq -z\leq 5-y\rightarrow y-5\leq z\leq y-4}$
• Now that have isolated the indicator for z, we can combine the entire integral for that indicator
• ${\displaystyle {\frac {1}{6}}\int _{-1}^{1}\left(z^{2}(y-z)\right)\cdot 1_{[y-5,y-4]}(z)\delta z={\frac {1}{6}}\int _{-1}^{1}\left(z^{2}y-z^{3}\right)\cdot 1_{[y-5,y-4]}(z)\delta z}$
• Finally, split the integral into the separate cases based on the remaining indicator function
• ${\displaystyle f_{X+Y}(z)={\frac {1}{6}}{\begin{cases}0&y<3\\\int _{-1}^{y-4}(z^{2}y-z^{3})\delta z&3\leq y<4\\\int _{y-5}^{y-4}(z^{2}y-z^{3})\delta z&4\leq y<5\\\int _{y-5}^{1}(z^{2}y-z^{3})\delta z&5\leq y<6\\0&y\geq 6\\\end{cases}}\qquad ={\begin{cases}0\\-{\frac {255}{4}}+43z+{\frac {z^{4}}{12}}-8z^{2}\\{\frac {369}{4}}-{\frac {122z}{3}}+{\frac {9z^{2}}{2}}\\156-83z-{\frac {z^{4}}{12}}+{\frac {25z^{2}}{2}}\\0\\\end{cases}}}$
• When ${\displaystyle y<3\,}$  the integral has no bounds since ${\displaystyle (y<3)-4<-1\,}$  so the upper bound would be less than ${\displaystyle -1\,}$  which would be ${\displaystyle 0\,}$ .
• When ${\displaystyle 3\leq y<4\,}$  the integral is bound between ${\displaystyle -1\,}$  and ${\displaystyle y-4\,}$ since ${\displaystyle y-4\,}$  will be at least ${\displaystyle -1\,}$  but less than ${\displaystyle 0\,}$
• As you can see there is a pattern here, it goes as follows:
• Given ${\displaystyle \int _{a}^{b}(\cdot )1_{[c,d]}\delta z\,}$  you will have ${\displaystyle {\begin{cases}y<(d-a=0)\\(d-a=0)\leq y<(c-a=0)\\(c-a=0)\leq y<(d-b=0)\\(d-b=0)\leq y<(c-b=0)\\y\geq (c-b=0)\\\end{cases}}}$